And you thought it was over?
Problem: Compute
,
and,
Here are some nice ones. Compute:
where is the Mobius function, is the divisor function, and counts the number of distinct primes dividing .
Edit: just to clarify, by I was referring, as usually denoted by this letter, to the function returning the sum of the divisors of , and not the number of divisors, which is also called the divisor function.
I think all of us would factorise the summation as this arithmetic function is multiplicative !
About the first problem , I remember it was first discovered by Ramanujan , he expanded the product :
As each factor provides only a prime but not a power of it , all the summands of the expansion should be square-free , and the signs of them exactly indicate the moebius function .
Therefore ,
But this time , absolute value is taken :